Question

One mole of an ideal monoatomic gas is taken along the path $ABCA$ as shown in the $PV$ diagram. The maximum temperature attained by the gas along the path $BC$ is given by :-

• A. $\frac{25}{8}\frac{P_0{V}_0}{R}$
• B. $\frac{25}{4}\frac{P_0{V}_0}{R}$
• C. $\frac{25}{16}\frac{P_0{V}_0}{R}$
• D. $\frac{5}{8}\frac{P_0{V}_0}{R}$

Answer 1

The correct option is A $\frac{25}{8}\frac{P_0{V}_0}{R}$

Temperature at A =$\frac{3P_{0}V}{nR}$

Temperature at b = $\frac{P_0\times 2V_0}{nR}$

Maximum temperature can be between B and C

P-V equation for process BC

$P-3P_0=\frac{P_0-3P_0}{2V_0-V_0}\times (V-V_0)$

$P-3P_0\frac{-2P_{0}V}{V_0}+2P_0$

$P=\frac{-2P_{0}V}{V_0}+2P_0$

Multiply by V

$PV=\frac{-2P_{0}V}{V_0}+5P_0$

$RT=-\frac{2P_0}{V_0}{V}^2+5P_{0}V$

Make $\frac{dT}{dV}=0$

This gives

$T=\frac{25P_0{V}_0}{8R}$